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In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that △ADE ≅ △BCE and hence, AEB is an isosceles triangle.

In the figure (i), CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that △ADE ≅ △BCE and hence, AEB is an isosceles triangle. Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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Answer

From figure,

∠ADE = ∠ADC + ∠CDE = 90° + 60° = 150°.

Similarly,

∠BCE = ∠BCD + ∠DCE = 90° + 60° = 150°.

⇒ ∠ADE = ∠BCE.

AD = BC (As sides of squares are equal)

DE = EC (As CDE is an equilateral triangle.)

∴ △ADE ≅ △BCE by SAS axiom.

We know that corresponding parts of congruent triangles are equal.

∴ AE = BE.

Hence, proved that AE = BE i.e. AEB is an isosceles triangle.

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